Variety of Nilpotent Matrices Let $k$ be an algebraically closed field and view $M_n(k)$ as $\mathbb{A}^{n^2}$.  $A\in M_n(k)$ is nilpotent if and only if $A^n=0$.  Since the equation $A^n=0$ is given by $n^2$ polynomial equations, it defines a variety in $\mathbb{A}^{n^2}$.  What is the dimension of this variety?  Is it irreducible?  If not, what are its irreducible components?
 A: It's dimension is $n(n-1)$ and it is indeed irreducible.
The variety you are describing is the nullcone of the Lie algebra $\mathfrak{gl}_n$.  It's known (with maybe some mild hypotheses that $\mathfrak{gl}_n$ satisfies) that the nullcone of a Lie algebra is irreducible and its dimension is twice the dimension of a maximal unipotent subalgebra.  In the case of $\mathfrak{gl}_n$ this will be the subalgebra $\mathfrak{u}_n$ of upper triangular matrices with zero's on the diagonal, which has dimension $\frac{1}{2}n(n-1)$.
Both these facts are hard, I can't explain the proof to you here.  If you want a reference, see Jantzens section of the Birkhauser Lie Theory book.
A: To prove irreducibility, note that any two regular nilpotent $n\times n$ matrices (i.e. nilpotent matrices whose minimal poly. is $X^n$) are conjugate (by putting them into Jordan normal form), and that any non-regular matrix can be written as a "limit" of regular ones (put it in Jordan form, than put an entry $t$ in any of the entries immediately above the diagonal that are zero, and let $t \to 0$).
From this we see that (a) the locus of regular nilpotent elements is irreducible (it is the image of $GL(n)$ under the conjugation action on any particular regular nilpotent), and (b) the whole nilpotent cone (another common synonym for the word nullcone in Jim's answer) is the irreducible, being the closure of the regular nilpotent elements.
A fairly simple computation shows that the centralizer of a regular nilpotent is $n$-dimensional (it is just the span of $I, A, \ldots, A^{n-1}$), and hence the image of a regular nilpotent under the conjugation action of $GL(n)$ is $n^2 - n$ dimensional.  Again considering (a) and (b) above, we see that the nilpotent cone is $n(n-1)$-dimensional.
Note also that the equation $char(A) = 0$ gives $n$ equations cutting out the nilpotent cone, so it is a complete intersection, and in particular Cohen--Macaulay (although it is not smooth, unless $n = 1$).  A little more argument builds on this to show it is regular in codimension one (i.e. its singular locus lies in codimension two), and hence that it is normal (by Serre's criterion for normality).
